poj 2184 Cow Exhibition

最新推荐文章于 2020-05-15 15:31:53 发布

原创最新推荐文章于 2020-05-15 15:31:53 发布 · 277 阅读

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通过深入分析每头牛的智能性和娱乐性，帮助Bessie组织展览，最大化展示牛群的综合价值，同时确保智能和娱乐性均为正数。

问题描述

"Fat and docile, big and dumb, they look so stupid, they aren't much
fun..."
- Cows with Guns by Dana Lyons

The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.

Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.

输入

* Line 1: A single integer N, the number of cows

* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.

输出

* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0.

样例输入

样例输出

8


 #include <iostream>
#include <cstdio>
using namespace std;

const int INF=1<<30;
int dp[200005],s[105],f[105];
int n;

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        
    
    for(int i=0;i<=200000;i++)
        dp[i]=-INF;
    dp[100000]=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&s[i],&f[i]);
    }
    for(int i=1;i<=n;i++)
    {
        if(s[i]<0&&f[i]<0)
            continue;
        if(s[i]>0)
        {
            for(int j=200000;j>=s[i];j--)
                if(dp[j-s[i]]>-INF)
                    dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
        }
        else
        {
            for(int j=s[i];j<=200000+s[i];j++)
                if(dp[j-s[i]]>-INF)
                    dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
        }
    }

    int ans=-INF;
    for(int i=100000;i<=200000;i++)
    {
        if(dp[i]>=0)
        ans=max(ans,dp[i]+i-100000);
    }
    printf("%d\n",ans);
    }
}